A202805 a(n) is the largest k in an n_nacci(k) sequence (Fibonacci(k) for n=2, tribonacci(k) for n=3, etc.) such that n_nacci(k) >= 2^(k-n-1).

6, 12, 25, 48, 94, 184, 363, 719, 1430, 2851, 5691, 11371, 22728, 45443, 90870, 181724, 363429, 726839, 1453658, 2907295, 5814566, 11629107

Offset: 2

Author

Frank M Jackson, Dec 24 2011

Keywords

• nonn
• more

Comments

From Frank M Jackson, Jul 02 2023: (Start)

Define the n_nacci sequence, basically row n in A092921, with an offset of 0, n_nacci(k) = 0 for 0 <= k <= n-2 and n_nacci(n-1) = 1. Thereafter, n_nacci(k) for k >= n continues as the sum of its previous n terms.

This means that n_nacci(k) = 2^(k-n) for n <= k <= 2n-1. In the limit as n tends to infinity the n_nacci sequence after an initial large set of zeros followed by 1 has successive terms of ascending powers of 2.

As the n-acci constants, (A001622, A058265, A086088, A103814,...) are smaller than 2, for each n_nacci sequence there is a largest k such that n_nacci(k) >= 2^(k-n-1). (End)

Examples

			For n=3, the tribonacci sequence is 0,0,1,1,2,4,7,...,149,274,504,... and the 13th term is 504 < 512 so a(n)=12 because 274 is greatest term >= 2^(12-3-1) = 256.
		

Crossrefs

Cf. A000045, A000073, A000078.

Programs

Maple

nAcci := proc(n,k)
    option remember ;
    if k <= n-2 then
        0;
    elif k = n-1 then
        1;
    else
        add( procname(n,i),i=k-n..k-1) ;
    end if;
end proc:
A202805 := proc(n)
    local k ;
    for k from n do
        if nAcci(n,k) < 2^(k-n-1) then
            return k-1;
        end if;
    end do:
end proc:
for n from 2 do
    print(n,A202805(n)) ;
end do: # R. J. Mathar, Mar 11 2024

Mathematica

fib[n_, m_] := (Block[{nacci}, (Do[nacci[g]=0, {g, 0, m - 2}];
nacci[m-1]=1;nacci[p_] := (nacci[p]=Sum[nacci[h], {h, p-m, p-1}]);nacci[n])]);
crossover[q_] := (Block[{$RecursionLimit=Infinity}, (k=0;While[fib[k+q+1, q]>=2^k, k++];k+q)]);
Table[crossover[j], {j, 2, 12}]

Python

def nacci(n): # generator of n_nacci terms
    window = [0]*(n-1) + [1]
    yield from window
    while True:
        an = sum(window)
        yield an
        window = window[1:] + [an]
def a(n):
    pow2 = 1
    for k, t in enumerate(nacci(n)):
        if k > n + 1: pow2 <<= 1
        if 0 < t < pow2: return k-1
print([a(n) for n in range(2, 12)]) # Michael S. Branicky, Jan 29 2025

Extensions

Edited by N. J. A. Sloane, May 20 2023

There seems to be an error in the Comment. See "History" tab. - N. J. A. Sloane, Jun 24 2023

Removed musing about what might define "complete" sequences. - R. J. Mathar, Mar 11 2024

a(17)-a(23) from Michael S. Branicky, Jan 29 2025